Question

5) resolva: x ao cubo - 13x ao quadrado + 36 = 0

Answer 1

Explicação passo-a-passo:

Equação biquadrática:

x⁴ — 13x² + 36 = 0

(x²)² — 13x² + 36 = 0

Seja : x² = t

t² — 13t + 36 = 0

Coeficientes:

a = 1

b = -13

c = 36

Bhaskara:

$\large\boxed{\boxed{{t_{(1,2)}=\frac{-b\pm\sqrt{b^2-4.a.c}}{2.a}}}}}}$

Jogando os coeficientes na fórmula ter-se-á:

$t_{(1,2)}=\frac{-(-13)\pm\sqrt{(-13)^2-4.1.36}}{2.1}$

$t_{(1,2)}=\frac{13\pm\sqrt{169-144}}{2}$

$t_{(1,2)}=\frac{13\pm\sqrt{25}}{2}$

$t_{(1,2)}=\frac{13\pm\:5}{2}$

$t_{1}=\frac{13+5}{2}=\frac{18}{2}={\color{blue}{9}}$

$t_{2}=\frac{13-5}{2}=\frac{8}{2}={\color{blue}{4}}$

Então para as raízes da nossa equação biquadrática:

teremos que:

$x_{(1,2)}=\pm\sqrt{t_{1}}$

$x_{1,2}=\pm\sqrt{9}$

$x_{1}={\color{blue}{+3}}$

$x_{2}={\color{blue}{-3}}$

$x_{(3,4)}=\pm\sqrt{t_{2}}$

$x_{(3,4)}=\pm\sqrt{4}$

$x_{3}={\color{blue}{+2}}$

$x_{4}={\color{blue}{-2}}$

Sol: { -2 , -3 , +2 , +3 }

Espero ter ajudado bastante!)

Answer 2

Resposta:

x⁴ - 13x² + 36 = 0 ➡ t = x².

t² - 13t + 36 = 0

-

x = -b±√{b²-4ac}/2a

t = -(- 13)±√{(- 13)²-4•1•36}/2•1

t = 13±√{169-144}/2

t = 13±√25/2

t = 13±5/2

t = 13+5/2 = 18/2 = 9

t = 13-5/2 = 8/2 = 4

-

t = 9 ➡ t = x² ➡ x² = 9

t = 4 ➡ t = x² ➡ x² = 4

-

x² = 9

x² = ± 3

x₁ = - 3

x₂ = 3

-

x² = 4

x² ± 2

x₁ = - 2

x₂ = 2

-

S {x₁ = - 3 , x₂ = 3 , x₃ = - 2 , x₄ = 2}